How to find sleeping process in Ubuntu?

Solution 1:

Try this:

ps o state,command axh | grep "^[SD]" | cut -b 3-

for listing commands of processes with an interruptable and uninterruptable sleep state.

• ps outputting only state and commands of all processes (ax) and h removes the header line.
• grep filters processes other than the two sleep states
• cut is used to remove the state output again.
• Optionally replace command with ucmd if you don't need the full name including all arguments.

This is probably suboptimal scripting here, but I couldn't find a quick way to have ps filtered for a specific state.

Solution 2:

You could grab the information from top, which can be run in batch mode (-b).

top -bn1 | awk 'NR > 7 && $8 ~ /S|D/ { print $12 }'

• -n1 top runs only once and exits.
• NR > 7 skips header.
• $8 ~ /S|D/ selects programs which are in state D or S.

Possible states are, from top(1):

      'D' = uninterruptible sleep
      'R' = running
      'S' = sleeping
      'T' = traced or stopped
      'Z' = zombie